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求不定积分∫xcos(x^2)dx
人气:379 ℃ 时间:2020-06-06 04:48:42
解答
∫xcos(x^2)dx
=∫cos(x^2)(xdx)
=∫cos(x^2)(d(x^2)/2)
=(1/2)∫cos(x^2)d(x^2)
=(1/2)sin(x^2)+C
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